Question

A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 8 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 48% chance, independently of each other.

A) Find the probability that overbooking occurs.
B) Find the probability that the flight has empty seats.

Answer 1

a) 32.04% probability that overbooking occurs.

b) 40.79% probability that the flight has empty seats.

Explanation:

For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. The probability of a passenger arriving is independent of other passengers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Our variable of interest are the 8 reservations that went for the passengers with a 48% probability of arriving.

This means that
`n = 8, p = 0.48`

A) Find the probability that overbooking occurs.

12 seats, 8 of which are already occupied. So overbooking occurs if more than 4 of the reservated arrive.

`P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 5) = C_(8,5).(0.48)^(5).(0.52)^(3) = 0.2006`

`P(X = 6) = C_(8,6).(0.48)^(6).(0.52)^(2) = 0.0926`

`P(X = 7) = C_(8,7).(0.48)^(7).(0.52)^(7) = 0.0244`

`P(X = 8) = C_(8,5).(0.48)^(8).(0.52)^(0) = 0.0028`

`P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2006 + 0.0926 + 0.0244 + 0.0028 = 0.3204`

32.04% probability that overbooking occurs.

B) Find the probability that the flight has empty seats.

Less than 4 of the booked passengers arrive.

To make it easier, i will use

`P(X < 4) = 1 - (P(X = 4) + P(X > 4))`

From a), P(X > 4) = 0.3204

`P(X = 4) = C_(8,4).(0.48)^(4).(0.52)^(4) = 0.2717`

`P(X < 4) = 1 - (P(X = 4) + P(X > 4)) = 1 - (0.2717 + 0.3204) = 1 - 0.5921 = 0.4079`

40.79% probability that the flight has empty seats.